Question

Suppose 83.0 mL. of dioxygen gas are produced by this reaction, at a temperature of 70.0 °C and pressure of exactly I atm. Calculate the mass of mercury(II) oxide that must have reacted. Round your answer to 3 significant digits.

Answer 1

Answer:

m = 0.3249 g

Explanation:

First, I'm assuming you have a reaction of mercury(II) oxyde descomposition. If this is the case, then the equation to use is the following:

HgO ---------> Hg + O2

Balancing the equation:

2HgO ----------> 2Hg + O2

This means that 2 moles of HgO reacts to produce 1 mole of O2, so, we first calculate the moles of O2, then, the moles of HgO and finally the mass:

We have the volume of O2, the pressure and temperature, so let's use the ideal gas equation:

PV = nRT

Solving for n:

n = PV/RT

R: 0.082 L atm / K mole

T = 70 + 273 = 343 K

V = 83 / 1000 = 0.083 L

Calculating n:

n = 1 * 0.083 / 0.082 * 343

n = 0.003 moles

as stated before, 2 moles of HgO reacts with 1 mole of oxygen so:

2 moles HgO = 1 moles O2

moles HgO = moles O2 / 2

moles HgO = 0.003 / 2 = 0.0015 moles

Finally, to calculate the mass:

m = n * MM

the molar mass of HgO is 216.59 g/mol, so replacing:

m = 0.0015 * 216.59

m = 0.3249 g